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d1i1m1o1n [39]
2 years ago
14

9-3 divided by 1/3+1

Mathematics
2 answers:
atroni [7]2 years ago
7 0

Answer:

i believe the answer is 1

Step-by-step explanation:

garri49 [273]2 years ago
3 0

Answer:

1

Step-by-step explanation:

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Write a linear function f with the values f (0) = 5 and f(2) = -3 please help
Gelneren [198K]

Answer:

f(x) = 2x - 5

Step-by-step explanation:

3 0
2 years ago
In ΔMNO, the measure of ∠O=90°, the measure of ∠M=46°, and NO = 96 feet. Find the length of OM to the nearest tenth of a foot.
GalinKa [24]

Answer:

92.7 ft

Step-by-step explanation:

6 0
2 years ago
A parabola has its focus at (1,2) and its directrix is y=-2. the equation of this parabola could be
Oksi-84 [34.3K]
<span>x^2/8 - x/4 + 1/8 = 0 A parabola is defined as the set of all points such that each point has the same distance from the focus and the directrix. Also the parabola's equation will be a quadratic equation of the form ax^2 + bx + c. So if we can determine 3 points on the parabola, we can use those points to calculate the desired equation. First, let's draw the shortest possible line from the focus to the directrix. The midpoint of that line will be a point on the desired parabola. Since the slope of the directrix is 0, the line will have the equation of x=1. This line segment will be from (1,2) to (1,-2) and the midpoint will be ((1+1)/2, (2 + -2)/2) = (2/2, 0/2) = (1,0). Now for the 2nd point, let's draw a line that's parallel to the directrix and passing through the focus. The equation of that line will be y=2. Any point on that line will have a distance of 4 from the directrix. So let's give it an x-coordinate value of (1+4) = 5. So another point for the parabola is (5,2). And finally, if we subtract 4 instead of adding 4 to the x coordinate, we can get a third point of 1-4 = -3. So that 3rd point is (-3,2). So we now have 3 points on the parabola. They are (1,0), (5,2), and (-3,2). Let's create some equations of the form ax^2 + bx + c = y and then substitute the known values into those equations. SO ax^2 + bx + c = y (1) a*1^2 + b*1 + c = 0 (2) a*5^2 + b*5 + c = 2 (3) a*(-3)^2 + b*(-3) + c = 2 Let's do the multiplication for those expressions. So (4) a + b + c = 0 (5) 25a + 5b + c = 2 (6) 9a - 3b + c = 2 Equations (5) and (6) above look interesting. Let's subtract (6) from (5). So 25a + 5b + c = 2 - 9a - 3b + c = 2 = 16a + 8b = 0 Now let's express a in terms of b. 16a + 8b = 0 16a = -8b a = -8b/16 (7) a = -b/2 Now let's substitute the value (-b/2) for a in expression (4) above. So a + b + c = 0 -b/2 + b + c = 0 And solve for c -b/2 + b + c = 0 b/2 + c = 0 (8) c = -b/2 So we know that a = -b/2 and c = -b/2. Let's substitute those values for a and c in equation (5) above and solve for b. 25a + 5b + c = 2 25(-b/2) + 5b - b/2 = 2 -25b/2 + 5b - b/2 = 2 2(-25b/2 + 5b - b/2) = 2*2 -25b + 10b - b = 4 -16b = 4 b = -4/16 b = -1/4 So we now know that b = -1/4. Using equations (7) and (8) above, let's calculate a and c. a = -b/2 = -(-1/4)/2 = 1/4 * 1/2 = 1/8 c = -b/2 = -(-1/4)/2 = 1/4 * 1/2 = 1/8 So both a and c are 1/8. So the equation for the parabola is x^2/8 - x/4 + 1/8 = 0 Let's test to make sure it works. First, let's use an x of 1. x^2/8 - x/4 + 1/8 = y 1^2/8 - 1/4 + 1/8 = y 1/8 - 1/4 + 1/8 = y 1/8 - 2/8 + 1/8 = y 0 = y And we get 0 as expected. Let's try x = 2 x^2/8 - x/4 + 1/8 = y 2^2/8 - 2/4 + 1/8 = y 4/8 - 1/2 + 1/8 = y 4/8 - 1/2 + 1/8 = y 1/2 - 1/2 + 1/8 = y 1/8 = y. Let's test if (2,1/8) is the same distance from both the focus and the directrix. The distance from the directrix is 1/8 - (-2) = 1/8 + 2 = 1/8 + 16/8 = 17/8 The distance from the focus is d = sqrt((2-1)^2 + (1/8-2)^2) d = sqrt(1^2 + -15/8^2) d = sqrt(1 + 225/64) d = sqrt(289/64) d = 17/8 And the distances match again. So we do have the correct equation of: x^2/8 - x/4 + 1/8 = 0</span>
4 0
3 years ago
What is 3x8+5x8-6x6-2x6
Schach [20]

Answer:

16

Step-by-step explanation:

group the first 2 terms into (8*8) and the second two into (8*6). this gets you 8*8-8*6 which can be grouped into 8*2 which is 16

3 0
2 years ago
Read 2 more answers
Find all polar coordinates of point p where p = (6, -pi/5)
Lelu [443]
Given a point in coordinates form (r,\theta), one can compute the cartesian form like this:
(r\cos\theta,r\sin\theta )
We have:
r=6,\theta=-\dfrac{\pi}{5}
We get the cartesian form then:
\left(6\cos-\dfrac{\pi}{5},6\sin-\dfrac{\pi}{5}\right)
4 0
3 years ago
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